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\author{五六七}
\title{常微分方程第6周作业（6.1）}
%\date{2025年10月14日}

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%\abstract{第6章第1节：例1、例子2、习题1(1)、2(1)。}

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\begin{enumerate}
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\item  例子6.1\#1
验证微分方程组
$$\frac{d}{dx} \begin{bmatrix} y_1(x) \\ y_2(x) \end{bmatrix} 
=\begin{bmatrix} \cos^2x & \frac{1}{2}\sin 2x -1 \\ \frac{1}{2}\sin 2x +1 & \sin ^2 x \end{bmatrix}
\begin{bmatrix} y_1(x) \\ y_2(x) \end{bmatrix}
$$
的通解为 
$$ \vec{y}:=\begin{bmatrix} y_1(x) \\ y_2(x) \end{bmatrix} 
=c_1\begin{bmatrix} e^x\cos x \\ e^x\sin x \end{bmatrix} 
+c_2\begin{bmatrix} -\sin x \\ \cos x \end{bmatrix} 
=: c_1\vec{y}_1 + c_2\vec{y}_2.
$$

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{\color{red}解答：
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\item  例子6.1\#2
在例子1的基础上，求解非齐次线性微分方程组的初值问题：
\begin{eqnarray*}
\frac{d}{dx} \begin{bmatrix} y_1(x) \\ y_2(x) \end{bmatrix} 
&=& \begin{bmatrix} \cos^2x & \frac{1}{2}\sin 2x -1 \\ \frac{1}{2}\sin 2x +1 & \sin ^2 x \end{bmatrix}
\begin{bmatrix} y_1(x) \\ y_2(x) \end{bmatrix} + \begin{bmatrix} \cos x \\ \sin x \end{bmatrix},  \\ 
\begin{bmatrix} y_1(0) \\ y_2(0) \end{bmatrix}&=& \begin{bmatrix} 0 \\ 1 \end{bmatrix}.
\end{eqnarray*}

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\item  习题6.1\#1(1)
求齐次线性微分方程组 $\frac{d \vec{y}}{dt}=A(t)\vec{y}$ 的通解，其中 $A(t)$ 分别为 
$A(t) = \begin{bmatrix} \frac{1}{t} & 0 \\ 0 & \frac{1}{t} \end{bmatrix}$. 

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\item  习题6.1\#2(1)
求非齐次线性微分方程组的初值问题，
\begin{eqnarray*}
\left\{\begin{array}{l}
\frac{dx}{dt} = 1-\frac{2}{t}x, \,\, \frac{dy}{dt} = x+y-1+\frac{2}{t}x, \,\, (t>0), \\ 
x(1)=\frac{1}{3},\,\, y(1)=-\frac{1}{3}. 
\end{array}\right. 
\end{eqnarray*}

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